# Service for Solving Linear Programming Problems

and other interesting typical problems
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## Example №1. Solving of a System of Linear Equations by the Gauss Elimination (One Solution)

This solution was made using the calculator presented on the site.
Please note that the coefficients will disappear which located in the "red" positions. 4 x1 + 2 x2 - 3 x3 = - 3 5 x1 + 3 x2 - 5 x3 = - 8 4 x1 + x2 + 5 x3 = 22
( 5 x1 + 4 x1 * ( -1) )
+ ( 3 x2 + 2 x2 * ( -1) )
+ ( -5 x3 + ( -3 x3) * ( -1) )
= -8 + ( -3) * ( -1)
This transformation will allow us to count without fractions for some time. 4 x1 + 2 x2 - 3 x3 = - 3 x1 + x2 - 2 x3 = - 5 4 x1 + x2 + 5 x3 = 22
The equation 2 and equation 1 are reversed. x1 + x2 - 2 x3 = - 5 4 x1 + 2 x2 - 3 x3 = - 3 4 x1 + x2 + 5 x3 = 22
( 4 x1 + x1 * ( -4) )
+ ( 2 x2 + x2 * ( -4) )
+ ( -3 x3 + ( -2 x3) * ( -4) )
= -3 + ( -5) * ( -4)
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 - 2 x2 + 5 x3 = 17 4 x1 + x2 + 5 x3 = 22
( 4 x1 + x1 * ( -4) )
+ ( x2 + x2 * ( -4) )
+ ( 5 x3 + ( -2 x3) * ( -4) )
= 22 + ( -5) * ( -4)
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 - 2 x2 + 5 x3 = 17 - 3 x2 + 13 x3 = 42
( -2 x2 + ( -3 x2) * ( -1) )
+ ( 5 x3 + 13 x3 * ( -1) )
= 17 + 42 * ( -1)
This transformation will allow us to count without fractions for some time. x1 + x2 - 2 x3 = - 5 x2 - 8 x3 = - 25 - 3 x2 + 13 x3 = 42
( -3 x2 + x2 * 3 )
+ ( 13 x3 + ( -8 x3) * 3 )
= 42 + ( -25) * 3
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 x2 - 8 x3 = - 25 - 11 x3 = - 33
We will find the variable x3 from equation 3 of the system.
- 11 x3 = - 33
x3 = 3
We will find the variable x2 from equation 2 of the system.
x2 - 8 x3 = - 25
x2 = - 25 + 8 x3
x2 = - 25 + 8 * ( 3 )
x2 = - 1
We will find the variable x1 from equation 1 of the system.
x1 + x2 - 2 x3 = - 5
x1 = - 5 - x2 + 2 x3
x1 = - 5 - ( - 1 ) + 2 * ( 3 )
x1 = 2
Result:
x1 = 2
x2 = - 1
x3 = 3